Find the distance between the point ${(-2, -6)}$ and the line $\enspace {y = x + 2}\thinspace$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
First, find the equation of the perpendicular line that passes through ${(-2, -6)}$ The slope of the blue line is ${1}$ , and its negative reciprocal is ${-1}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -x + b}\thinspace$ We can plug our point, ${(-2, -6)}$ , into this equation to solve for ${b}$ , the y-intercept. $-6 = {-}(-2) + {b}$ $-6 = 2 + {b}$ $-6 - 2 = {b} = -8$ The equation of the perpendicular line is $\enspace {y = -x - 8}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(-5, -3)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(-2, -6)}$ and ${(-5, -3)}$ gives us: $\sqrt{( {-2} - {-5} )^2 + ( {-6} - {-3} )^2}$ $= \sqrt{( 3 )^2 + ( -3 )^2} = \sqrt{18} = 3\sqrt{2}$ The distance between the point ${(-2, -6)}$ and the line $\thinspace {y = x + 2}\enspace$ is $\thinspace3\sqrt{2}$.